(a) how much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from −20.0ºc to 130ºc, including the energy needed for phase changes? (b) how much time is required for each stage, assuming a constant 20.0 kj/s rate of heat transfer?

a)Q=167.5 KJ

b)t=8.375 s

Explanation:

a)Given that

mass m= 0.2 kg

We know that

Cp for ice is 2.1 KJ/Kg.K

Cp for water is 4.718 KJ/Kg.K

Cp for steam is 1.9 KJ/Kg.K

Latent heat of ice=336 KJ/kg

First ice will take sensible heat and increases temperature from −20.0ºC to 0ºC after that it will  transfer phase from solid to liquid at constant temperature and will take latent heat after that water will take sensible heat and  increases temperature from 0ºC to 100ºC. Above 100ºC water will convert in to vapor and vapor will take sensible heat  increases temperature from 100ºC to 130ºC .

Sensible heat = m CpΔT KJ

Latent heat = m L.H.  KJ

So total heat

Q=0.2(2.1 x 20 + 336 + 4.178 x 100 + 1.9 x 130) KJ

Q=167.5 KJ

b)

Heat transfer = 20 KJ/s

Lets time taken is t

20 t =167.5

t=8.375 s

terd muffin is the correct answer